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A survey claims that the average cost of a hotel room in tulsa, oklahoma, is less than $49.21. ... Your question has been answered Let us know if you got a helpful answer. Rate this answer Question: A survey claims that the average cost of a hotel room in Tulsa, Oklahoma, is less than $49.21. In... See this question in the app A survey claims that the average cost of a hotel room in Tulsa, Oklahoma, is less than $49.21. In order to test this claim, a researcher selects a sample of 26 hotel rooms and finds the average cost equal to $47.37 with a sample standard deviation of $6.42. Assume the population random variable is normal. At alpha=.1. What is the p-value of the test.

Answer #1

The average cost of a hotel room in NYC is $168 per night. A
random sample of 25 hotel rooms resulted in X = $172.50 and s =
$15.40. Test the claim at the 0.05 level.

A state executive claims that the average cost of tuition at
public colleges is $5700. A researcher wishes to test the claim
that this cost is greater than $5700. She selects a random sample
of 36 public colleges and finds the mean to be $5950.The population
standard deviation is $659. Is there enough evidence to support the
claim that the cost of tuition is greater than $5700 at α = 0.05
?

Question 1
The average cost per night of a hotel room in New York City is
$273 (SmartMoney, March 2009). Assume this estimate is based on a
sample of 45 hotels and that the sample standard deviation is
$65.
a. With 95% confidence, what is the margin of error?
b. What is the 95% confidence interval estimate of the population
mean?
c. Two years ago the average cost of a hotel room in New York
City was $229. Discuss the...

1.
(CO 7) A travel analyst claims that the mean room rates at a
three-star hotel in Chicago is greater than $152. In a random
sample of 36 three-star hotel rooms in Chicago, the mean room rate
is $163 with a standard deviation of $41. At α=0.10, what type of
test is this and can you support the analyst’s claim using the
p-value?
Claim is the alternative, fail to reject the null as p-value
(0.054) is less than alpha (0.10),...

A hospital claims that its emergency room has an average wait
time of 2 hours (120 minutes). After a visit to their ER when we
had to wait for what seemed like an eternity, we decide to test the
hospital's claim because we think their claim of an average of two
hours is way too low. We survey people as they exit the ER. From
our sample of 58 people, we get a mean wait time of 128 minutes
with...

A researcher wished to compare the average daily hotel room
rates between San Francisco and Los Angeles. The researcher
obtained an SRS of 15 hotels in downtown San Francisco and found
the sample mean x1=$156 , with a standard deviation s1= $15 . The
researcher also obtained an independent SRS of 10 hotels in
downtown Los Angeles and found the sample mean x2= $143, with a
standard deviation s2=$10.
Let 1 and 2 represent the mean cost of the populations...

Many people claimed that the average cost of living for grocery
expenses for family with one or two children in the Washington D.C
metropolitan is exceeded $1,200 per month. In a sample of 100
family survey undertaken in the Washington D.C metropolitan area in
July 2019. The average cost and standard deviation of living for
grocery expenses for family with one or two children in the
Washington D.C metropolitan are found to be $1,380 and $190,
respectively. Do these data...

A marketing manager claims that the average 18-25 year old
watches Netflix less than 26 hours per week. She collects data on
25 individuals' in the age range and interviews them about their
Netflix habits. She finds that the mean number of hours that the
sample spent watching Netflix was 22.4 hours. If the population
standard deviation is known to be eight hours, can we conclude at
the 1% significance level that she is right? State the null and
alternative...

3. Insurance company A claims that its customers pay less for
car insurance, on average, than customers of its competitor,
Company B. You wonder if this is true, so you decide to compare the
average monthly costs of similar insurance policy from the two
companies. For a random sample of 9 people who buy insurance from
company A, the mean cost is $152 per month with a sample standard
deviation of $17. For 11 randomly selected customers of company B,...

A "sleep habits" survey answered by 23 randomly selected New
Yorkers contained the question "How much sleep do you get per
night?" The sample average was 7.75 hours, with a corresponding
sample standard deviation of 0.8 hours. Conduct a hypothesis test
to see if there is evidence that New Yorkers (who live, after all,
in "the city that never sleeps") on average get less than 8 hours
of sleep per night, using α=0.05.
a. The t test statistic is:
Round...

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